NNG progress
This commit is contained in:
Jon Eugster
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import NNG.Metadata
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import NNG.MyNat.Addition
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import NNG.MyNat.Multiplication
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import NNG.Levels.Addition
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Game "NNG"
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World "Multiplication"
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Level 1
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Title ""
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Title "zero_mul"
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open MyNat
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Introduction
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"
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As a side note, the lemmas about addition are still available in your inventory
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in the lemma tab \"Add\" while the new lemmas about multiplication appear in the
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tab \"Mul\".
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We are given `mul_zero`, and the first thing to prove is `zero_mul`.
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Like `zero_add`, we of course prove it by induction.
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"
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Statement
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""
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: true := by
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trivial
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Statement MyNat.zero_mul
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"For all natural numbers $m$, we have $ 0 \\cdot m = 0$."
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(m : ℕ) : 0 * m = 0 := by
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induction m
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rw [mul_zero]
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rfl
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rw [mul_succ]
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rw [n_ih]
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rw [add_zero]
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rfl
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Conclusion
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"
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"
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NewTactic simp
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NewLemma MyNat.mul_zero MyNat.mul_succ
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NewDefinition Mul
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LemmaTab "Mul"
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@@ -1,24 +1,31 @@
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import NNG.Metadata
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import NNG.MyNat.Addition
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import NNG.Levels.Multiplication.Level_1
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Game "NNG"
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World "Multiplication"
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Level 2
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Title ""
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Title "mul_one"
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open MyNat
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Introduction
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"
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In this level we'll need to use
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* `one_eq_succ_zero : 1 = succ 0 `
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which was mentioned back in Addition World (see \"Add\" tab in your inventory) and
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which will be a useful thing to rewrite right now, as we
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begin to prove a couple of lemmas about how `1` behaves
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with respect to multiplication.
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"
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Statement
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""
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: true := by
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trivial
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Statement MyNat.mul_one
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"For any natural number $m$, we have $ m \\cdot 1 = m$."
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(m : ℕ) : m * 1 = m := by
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rw [one_eq_succ_zero]
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rw [mul_succ]
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rw [mul_zero]
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rw [zero_add]
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rfl
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Conclusion
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"
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"
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LemmaTab "Mul"
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@@ -1,24 +1,38 @@
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import NNG.Metadata
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import NNG.MyNat.Addition
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import NNG.Levels.Multiplication.Level_2
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Game "NNG"
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World "Multiplication"
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Level 3
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Title ""
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Title "one_mul"
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open MyNat
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Introduction
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"
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These proofs from addition world might be useful here:
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* `one_eq_succ_zero : 1 = succ 0`
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* `succ_eq_add_one a : succ a = a + 1`
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We just proved `mul_one`, now let's prove `one_mul`.
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Then we will have proved, in fancy terms,
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that 1 is a \"left and right identity\"
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for multiplication (just like we showed that
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0 is a left and right identity for addition
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with `add_zero` and `zero_add`).
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"
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Statement
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""
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: true := by
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trivial
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Statement MyNat.one_mul
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"For any natural number $m$, we have $ 1 \\cdot m = m$."
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(m : ℕ): 1 * m = m := by
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induction m with d hd
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· rw [mul_zero]
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rfl
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· rw [mul_succ]
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rw [hd]
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rw [succ_eq_add_one]
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rfl
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Conclusion
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"
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LemmaTab "Mul"
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"
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Conclusion ""
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@@ -1,24 +1,44 @@
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import NNG.Metadata
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import NNG.MyNat.Addition
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import NNG.Levels.Multiplication.Level_3
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Game "NNG"
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World "Multiplication"
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Level 4
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Title ""
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Title "mul_add"
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open MyNat
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Introduction
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"
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Where are we going? Well we want to prove `mul_comm`
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and `mul_assoc`, i.e. that `a * b = b * a` and
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`(a * b) * c = a * (b * c)`. But we *also* want to
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establish the way multiplication interacts with addition,
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i.e. we want to prove that we can \"expand out the brackets\"
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and show `a * (b + c) = (a * b) + (a * c)`.
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The technical term for this is \"left distributivity of
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multiplication over addition\" (there is also right distributivity,
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which we'll get to later).
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Note the name of this proof -- `mul_add`. And note the left
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hand side -- `a * (b + c)`, a multiplication and then an addition.
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I think `mul_add` is much easier to remember than \"`left_distrib`\",
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an alternative name for the proof of this lemma.
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"
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Statement
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""
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: true := by
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trivial
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Statement MyNat.mul_add
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"Multiplication is distributive over addition.
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In other words, for all natural numbers $a$, $b$ and $t$, we have
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$t(a + b) = ta + tb$."
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(t a b : ℕ) : t * (a + b) = t * a + t * b := by
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induction b with d hd
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· rw [add_zero, mul_zero, add_zero]
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rfl
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· rw [add_succ]
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rw [mul_succ]
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rw [hd]
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rw [mul_succ]
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rw [add_assoc]
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rfl
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Conclusion
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"
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LemmaTab "Mul"
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"
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@@ -1,22 +1,41 @@
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import NNG.Metadata
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import NNG.MyNat.Addition
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import NNG.Levels.Multiplication.Level_4
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Game "NNG"
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World "Multiplication"
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Level 5
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Title ""
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Title "mul_assoc"
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open MyNat
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Introduction
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"
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We now have enough to prove that multiplication is associative.
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## Random tactic hint
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You can do `repeat rw [mul_succ]` to repeat a tactic as often as possible.
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"
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Statement
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""
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: true := by
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trivial
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Statement MyNat.mul_assoc
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"Multiplication is associative.
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In other words, for all natural numbers $a$, $b$ and $c$, we have
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$(ab)c = a(bc)$."
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(a b c : ℕ) : (a * b) * c = a * (b * c) := by
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induction c with d hd
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· repeat rw [mul_zero]
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rfl
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· rw [mul_succ]
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rw [mul_succ]
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rw [hd]
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rw [mul_add]
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rfl
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NewTactic «repeat»
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LemmaTab "Mul"
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-- TODO: old version introduced `rwa` here, but it would need to be modified
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-- as our `rw` does not call `rfl` at the end.
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Conclusion
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"
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@@ -1,22 +1,50 @@
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import NNG.Metadata
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import NNG.MyNat.Addition
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import NNG.Levels.Multiplication.Level_5
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Game "NNG"
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World "Multiplication"
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Level 6
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Title ""
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Title "succ_mul"
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open MyNat
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Introduction
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"
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We now begin our journey to `mul_comm`, the proof that `a * b = b * a`.
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We'll get there in level 8. Until we're there, it is frustrating
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but true that we cannot assume commutativity. We have `mul_succ`
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but we're going to need `succ_mul` (guess what it says -- maybe you
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are getting the hang of Lean's naming conventions).
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Remember also that we have tools like
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* `add_right_comm a b c : a + b + c = a + c + b`
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These things are the tools we need to slowly build up the results
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which we will need to do mathematics \"normally\".
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We also now have access to Lean's `simp` tactic,
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which will solve any goal which just needs a bunch
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of rewrites of `add_assoc` and `add_comm`. Use if
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you're getting lazy!
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"
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Statement
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""
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: true := by
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trivial
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Statement MyNat.succ_mul
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"For all natural numbers $a$ and $b$, we have
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$\\operatorname{succ}(a) \\cdot b = ab + b$."
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(a b : ℕ) : succ a * b = a * b + b := by
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induction b with d hd
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· rw [mul_zero]
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rw [mul_zero]
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rw [add_zero]
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rfl
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· rw [mul_succ]
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rw [mul_succ]
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rw [hd]
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rw [add_succ]
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rw [add_succ]
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rw [add_right_comm]
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rfl
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LemmaTab "Mul"
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Conclusion
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"
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@@ -1,22 +1,44 @@
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import NNG.Metadata
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import NNG.MyNat.Addition
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import NNG.Levels.Multiplication.Level_6
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Game "NNG"
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World "Multiplication"
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Level 7
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Title ""
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Title "add_mul"
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open MyNat
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Introduction
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"
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We proved `mul_add` already, but because we don't have commutativity yet
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we also need to prove `add_mul`. We have a bunch of tools now, so this won't
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be too hard. You know what -- you can do this one by induction on any of
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the variables. Try them all! Which works best? If you can't face
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doing all the commutativity and associativity, remember the high-powered
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`simp` tactic mentioned at the bottom of Addition World level 6,
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which will solve any puzzle which needs only commutativity
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and associativity. If your goal looks like `a + (b + c) = c + b + a` or something,
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don't mess around doing it explicitly with `add_comm` and `add_assoc`,
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just try `simp`.
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"
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Statement
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""
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: true := by
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trivial
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Statement MyNat.add_mul
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"Addition is distributive over multiplication.
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In other words, for all natural numbers $a$, $b$ and $t$, we have
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$(a + b) \times t = at + bt$."
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(a b t : ℕ) : (a + b) * t = a * t + b * t := by
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induction b with d hd
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· rw [zero_mul]
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rw [add_zero]
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rw [add_zero]
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rfl
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· rw [add_succ]
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rw [succ_mul]
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rw [hd]
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rw [succ_mul]
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rw [add_assoc]
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rfl
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LemmaTab "Mul"
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Conclusion
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"
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@@ -1,24 +1,46 @@
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import NNG.Metadata
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import NNG.MyNat.Addition
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import NNG.Levels.Multiplication.Level_7
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Game "NNG"
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World "Multiplication"
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Level 8
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Title ""
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Title "mul_comm"
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open MyNat
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Introduction
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"
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Finally, the boss level of multiplication world. But (assuming you
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didn't cheat) you are well-prepared for it -- you have `zero_mul`
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and `mul_zero`, as well as `succ_mul` and `mul_succ`. After this
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level you can of course throw away one of each pair if you like,
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but I would recommend you hold on to them, sometimes it's convenient
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to have exactly the right tools to do a job.
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"
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Statement
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""
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: true := by
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trivial
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Statement MyNat.mul_comm
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"Multiplication is commutative."
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(a b : ℕ) : a * b = b * a := by
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induction b with d hd
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· rw [zero_mul]
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rw [mul_zero]
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rfl
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· rw [succ_mul]
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rw [← hd]
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rw [mul_succ]
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rfl
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LemmaTab "Mul"
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Conclusion
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"
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You've now proved that the natural numbers are a commutative semiring!
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That's the last collectible in Multiplication World.
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* `CommSemiring ℕ`
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But don't leave multiplication just yet -- prove `mul_left_comm`, the last
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level of the world, and then we can beef up the power of `simp`.
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"
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-- TODO: collectible
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-- instance mynat.comm_semiring : comm_semiring mynat := by structure_helper
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@@ -1,24 +1,57 @@
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import NNG.Metadata
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import NNG.MyNat.Addition
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import NNG.Levels.Multiplication.Level_8
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Game "NNG"
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World "Multiplication"
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Level 9
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Title ""
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Title "mul_left_comm"
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open MyNat
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Introduction
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"
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You are equipped with
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* `mul_assoc (a b c : ℕ) : (a * b) * c = a * (b * c)`
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* `mul_comm (a b : ℕ) : a * b = b * a`
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Re-read the docs for `rw` so you know all the tricks.
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You can see them in your inventory on the right.
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"
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Statement
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""
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: true := by
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trivial
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Statement MyNat.mul_left_comm
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"For all natural numbers $a$ $b$ and $c$, we have $a(bc)=b(ac)$."
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(a b c : ℕ) : a * (b * c) = b * (a * c) := by
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rw [← mul_assoc]
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rw [mul_comm a]
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rw [mul_assoc]
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rfl
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LemmaTab "Mul"
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-- TODO: make simp work:
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-- attribute [simp] mul_assoc mul_comm mul_left_comm
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Conclusion
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"
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And now I whisper a magic incantation
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```
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attribute [simp] mul_assoc mul_comm mul_left_comm
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```
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and all of a sudden Lean can automatically do levels which are
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very boring for a human, for example
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```
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example (a b c d e : ℕ) :
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(((a * b) * c) * d) * e = (c * ((b * e) * a)) * d := by
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simp
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```
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If you feel like attempting Advanced Multiplication world
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you'll have to do Function World and the Proposition Worlds first.
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These worlds assume a certain amount of mathematical maturity
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(perhaps 1st year undergraduate level).
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Your other possibility is Power World, with the \"final boss\".
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"
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Reference in New Issue
Block a user